## neverforgetvivistee 4 years ago solving rational equations: x+2 x - 1 4x+2 ---- = ---- - ----- x x x^2-3x I know how to make them all have a common denominator x(x-3) but when it comes to solving I get x^2+5x+6=x^2-4x-3x-3x-2 which then comes up to be x^2+3x-11 This isn't right. What did I do wrong?

1. satellite73

maybe it would be easier to multiply both sides by \[x(x-3)\]

2. neverforgetvivistee

i did

3. neverforgetvivistee

but i can't figure out what i did wrong because i get the wrong answer... when i multiply x+2 and x-3 i get x^2+5x+5 when i multiply x-1 and x-3 i get x^2-4x-3

4. satellite73

you get \[(x+2)(x-3)=(x-1)(x-3)-4x+2\] i htink

5. satellite73

hold on \[(x+2)(x-3)=x^2-x-6\] ,maybe that is the mistake

6. satellite73

also \[(x-1)(x-3)=x^2-4x+3\]

7. neverforgetvivistee

oh!!!!!! i put (x+2) and postive (x+3)!!

8. neverforgetvivistee

that was a waste of time i was looking just at how i solved it and not how i multiplied it. Thanks

9. satellite73

\[(x+2)(x-3)=(x-1)(x-3)-4x+2\] \[x^2-x-6=x^2-4x+3-4x+2\] \[x^2-x-6=x^2-8x+5\] etc

10. satellite73

yw

11. neverforgetvivistee

when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative? for examples x-1 4x+2 ---- - ----- x(x-3) x(x-3) would you make it into a plus sign, and then make 4x and 2 negative? -4x-2

12. neverforgetvivistee

bump